practical 6 colony forming units calculation help
I anticipate that a lot of you will struggle in your next practical with calculating the number of colony forming units per gram of mushroom compost. Here is a worked example for you...
First of all, what is a colony forming unit? Well, when you look at your mushroom compost plates, you should hopefully see nice separate colonies of bacteria. We assume that each colony has grown up from a single bacterial cell. So a colony forming unit is a live bacterial cell that could create a colony on your agar plate. Therefore, if we count the number of colonies on an agar plate, this is an approximation of the number of live bacterial cells that you originally plated onto the plate.
But rather than just stating how many bacteria you had on your plate originally (which we assume is approximately equal to the number of colonies on the plate), you are asked to calculate how many colony forming units (i.e. individual alive bacteria) were in 1 g of your original mushroom compost.
To do this you have to remember what you did to your mushroom compost last week.... here is a quick summary:
NOTE: the dilutions you plated out will be different depending on which treatment you were using - if you were pair 6 (i.e. incubating your compost sample with Novobiocin) you probably plated out your undiluted sample and your 10-1 and 10-2 samples. If you were in pair 1, 2 or 3, you probably made and plated out 10-2, 10-3 and 10-4 dilutions.
Here is how I would go from the number of colonies on a plate to the number of colony forming units per gram of original compost sample:
1. First of all I would choose to count the number of colonies on the plate that gave the most countable number of colonies - as a rough guide you should choose a plate that has between 20 - 200 colonies, so that you have enough colonies to count to reduce errors but not so many to count that your colonies are overlapping each other. In my example above, I would choose to count the plate on which the 10-2 sample was plated. So in this case there are 36 colonies on the plate.
2. 36 colonies on the plate will roughly mean that 36 live bacterial cells were plated onto that plate. That means that in your 0.1 ml of 10-2 sample there were 36 bacteria.
3. Ok, so how many bacteria would there be in 0.1 ml of undiluted sample? Well, to get to the 10-2 dilution, you diluted your original sample by 10x twice (see my diagram above). So to undo this dilution you need to do 36 x 10 x 10 = 3600. This is the same as saying you diluted your original sample by 100x, so you could also calculate how many cells are in 0.1ml of undiluted sample by doing 36 x 100 = 3600.
4. Marvellous, now you know that you had 3600 cells in 0.1ml of undiluted sample. But you want to know how many colony forming units (i.e. live cells) are in 1 g of your compost.....
5. So, you have to first of all find out how many cells were in your original 10 mls of sample (remember, you suspended your compost in 10 mls of buffer). You can figure this out by doing 3600 x 10 = 36000 -> this is now the number of cells in 1ml of sample (since 0.1 ml fits into 1 ml ten times) and then you can do 36000 x 10 = 360000 -> this is now the number of cells in 10ml of your sample (since 1 ml fits into 10 mls ten times).
6. Great, so you now know that you had 360000 cells in your original 10 mls of mushroom compost suspension.
7. Now you need to remember that you were given 0.5 g of mushroom compost, and you need to find out how many cells there were in 1 g of compost. Seeing as you know that there were 360000 cells in 10 mls of your original sample, this means you know that there were 360000 cells in 0.5 g of compost. So, to find out how many cells there were in 1 g of compost you just have to do 360000 x 2 = 720000
So, the answer is 720000 cells per gram of compost, or to be more precise, 720000 colony forming units per gram of compost.
This could also be written as 7.2 x 105 Cfu.g-1
Hope that helps
First of all, what is a colony forming unit? Well, when you look at your mushroom compost plates, you should hopefully see nice separate colonies of bacteria. We assume that each colony has grown up from a single bacterial cell. So a colony forming unit is a live bacterial cell that could create a colony on your agar plate. Therefore, if we count the number of colonies on an agar plate, this is an approximation of the number of live bacterial cells that you originally plated onto the plate.
But rather than just stating how many bacteria you had on your plate originally (which we assume is approximately equal to the number of colonies on the plate), you are asked to calculate how many colony forming units (i.e. individual alive bacteria) were in 1 g of your original mushroom compost.
To do this you have to remember what you did to your mushroom compost last week.... here is a quick summary:
NOTE: the dilutions you plated out will be different depending on which treatment you were using - if you were pair 6 (i.e. incubating your compost sample with Novobiocin) you probably plated out your undiluted sample and your 10-1 and 10-2 samples. If you were in pair 1, 2 or 3, you probably made and plated out 10-2, 10-3 and 10-4 dilutions.
Here is how I would go from the number of colonies on a plate to the number of colony forming units per gram of original compost sample:
1. First of all I would choose to count the number of colonies on the plate that gave the most countable number of colonies - as a rough guide you should choose a plate that has between 20 - 200 colonies, so that you have enough colonies to count to reduce errors but not so many to count that your colonies are overlapping each other. In my example above, I would choose to count the plate on which the 10-2 sample was plated. So in this case there are 36 colonies on the plate.
2. 36 colonies on the plate will roughly mean that 36 live bacterial cells were plated onto that plate. That means that in your 0.1 ml of 10-2 sample there were 36 bacteria.
3. Ok, so how many bacteria would there be in 0.1 ml of undiluted sample? Well, to get to the 10-2 dilution, you diluted your original sample by 10x twice (see my diagram above). So to undo this dilution you need to do 36 x 10 x 10 = 3600. This is the same as saying you diluted your original sample by 100x, so you could also calculate how many cells are in 0.1ml of undiluted sample by doing 36 x 100 = 3600.
4. Marvellous, now you know that you had 3600 cells in 0.1ml of undiluted sample. But you want to know how many colony forming units (i.e. live cells) are in 1 g of your compost.....
5. So, you have to first of all find out how many cells were in your original 10 mls of sample (remember, you suspended your compost in 10 mls of buffer). You can figure this out by doing 3600 x 10 = 36000 -> this is now the number of cells in 1ml of sample (since 0.1 ml fits into 1 ml ten times) and then you can do 36000 x 10 = 360000 -> this is now the number of cells in 10ml of your sample (since 1 ml fits into 10 mls ten times).
6. Great, so you now know that you had 360000 cells in your original 10 mls of mushroom compost suspension.
7. Now you need to remember that you were given 0.5 g of mushroom compost, and you need to find out how many cells there were in 1 g of compost. Seeing as you know that there were 360000 cells in 10 mls of your original sample, this means you know that there were 360000 cells in 0.5 g of compost. So, to find out how many cells there were in 1 g of compost you just have to do 360000 x 2 = 720000
So, the answer is 720000 cells per gram of compost, or to be more precise, 720000 colony forming units per gram of compost.
This could also be written as 7.2 x 105 Cfu.g-1
Hope that helps
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